Insights Fermat's Last Theorem

[martinbn]

One tip for anyone on this thread. The result of what we are doing with integers and integer factorisation will remove the ability to use any form of modular exponentiation, or elliptic curve prime field for encryption. It is very likely the outcome of our work will result in it not being possible to use AES or anything like it for encryption or one way hashing (i.e. if I know what I need to hash to, I can pick any of the numbers that hash to it easily). We are not seeing a mathematical way to encrypt or hash one way after we publish.

This will result in the collapse of bitcoin and other coins in likelihood since proof-of-work is trivialised.

We are not the only party that has figured this out. It is likely actors are going to rug pull bitcoin for sure in the near future.

Example: RSA 100. if we take (d+n)(d+n) - (x+n)(x+n) = c (diff of two squares, d is the root of c)

See that 2a/x is almost exactly equal to x/n, it is only the remainder that fills in the blanks.

There is a class of integers like this. Essential the ratio of the remainder of dividing x by n is in proportion exactly to the ratio of dividing 2a by x. And 2d is 2a + 2x. So for RSA100 we can divided 2d by n. We don't know what n is so we use the number 1 to 10,000 as the amount of n in 2d. We get to 5424. Easy compared to the general number field sieve. At this number we have a chunk of the start of numbers...

a must start with 379... b must start with 400... x must start with 1054 and n must start with 1438... and theta must start with 72.655 (where theta is 2a/x and x/n rememeber that for all c, that if 72.655 x = 2a, then 74.655.. will equal 2d, since (73.655... x 73.655... x n ) squared equals c.

What is so good about that?

The values of x, n, and theta are not random and only possess certain characteristics that are related to the golden ratio, the squre root of two (you can see that in 1438... for n) and the decimal unit 1.

Because these numbers are fixed, it is very simple and very quick to figure out the expansion of the natural log of c, and which are the squares in the exponential function definition that are the ones that give the exact result for c, half of that exponent is obviously the square root of c, and if we know the five contributors in the natural log of c (five dimensions is always the max needed for any problem), then like finding the place of Pi without those in front, we spigot the solution. Works best for the product of two primes.

There are a fixed number of theses types of integer and it's like whether the square is odd and the remainder is even, the other way round. If n is smaller or larger than x and how does that affect theta. And the 'balance number' for that type of integer. In other words, what is the constant that all our variable must equal. This is what maths has been missing for integer factorisation. That pattern that is fixed. The larger the numbers are the easier it is to find.

What we are seeing evidence of is that P = NP. If a problem statement is complete, it is also the solution.

One tip for you too. If you talk about something you have no clue, the result is pure nonsens.

 

[noelbcornerstone]

All this is nonsense!!

If what yiu wrote made sense it would work for n=2 as well.

Suppose there is a solution to a^2+b^2=c^2, writw it as

1) a+b=c

Or

2) a+b=c+k

Multiply by c and run the same argument. Thus you have proven that there no pythagorian triples.

a^2 + b^2 = c^2 does have solutions, infinite solutions, that's my point. You would be able to divide both sides of any solution by c, so in any solution dividing by c must remove an indice from the left hand side. In the two cases there are only solutions to 1, and 2.

a + b = c

let's rewrite this as a^2 + b^2 = c^2 + k, where k = 0

I want to raise the indice of c by one and a and b by 1.

We know a + b cannot equal c, since the square root of a^2 + b^2 cannot be a + b

We know a + b = c + k must have a solution then if any solution exists and we state that there is a d^2 + e^2 - k = c^2

If we divide by c, we need to remove -k, and d^2 + e^2 must be divisible by c to give a + b. Infinite solutions to that, especially if k = 0 :)

I think because I didn't write out the examples, now you can see how d and e are employed, and you don't think about the initial a and b. In other words, you are not reading or understanding that the equations:

1. a^2 + b^2 = c^2

and

2) a^2 + b^2 = c^2 + k

represent all integers and any solution that can exist for any powers for FLT

What was confusing is I made it look like I was keep the original integers for a and b and this was a bit lazy, yet depends how you do it, if you understand those two equations represent all integers and all possible solutions, it's kinda obvious really why looking at ALL the indices is a pretty dumb approach when they are all represented by two simple equations. ALL possible solutions are represented above.

 

[fresh_42]

If you cannot understand that 1) and 2) cover all integers, can you please explain why?

Because $k=c^2-a^2-b^2$ is so arbitrary that it is useless, furthermore, division by $c$ leaves the realm of integers. And what does it have to do with $a^3+b^3=c^3$?

Besides that, I have no idea what you are talking about. Please write down the case $n=4$ step by step, and you will see where the problems are; or I can tell you. But this general handwavy form is almost impossible to deal with.

 

[noelbcornerstone]

a^2 + b^2 = c^2 does have solutions, infinite solutions, that's my point. You would be able to divide both sides of any solution by c, so in any solution dividing by c must remove an indice from the left hand side. In the two cases there are only solutions to 1, and 2.

a + b = c

let's rewrite this as a^2 + b^2 = c^2 + k, where k = 0

I want to raise the indice of c by one and a and b by 1.

We know a + b cannot equal c, since the square root of a^2 + b^2 cannot be a + b

We know a + b = c + k must have a solution then if any solution exists and we state that there is a d^2 + e^2 - k = c^2

If we divide by c, we need to remove -k, and d^2 + e^2 must be divisible by c to give a + b. Infinite solutions to that, especially if k = 0 :)

I think because I didn't write out the examples, now you can see how d and e are employed, and you don't think about the initial a and b. In other words, you are not reading or understanding that the equations:

1. a^2 + b^2 = c^2

and

2) a^2 + b^2 = c^2 + k

represent all integers and any solution that can exist for any powers for FLT

What was confusing is I made it look like I was keep the original integers for a and b and this was a bit lazy, yet depends how you do it, if you understand those two equations represent all integers and all possible solutions, it's kinda obvious really why looking at ALL the indices is a pretty dumb approach when they are all represented by two simple equations. ALL possible solutions are represented above.

The REALLY important part above should be clarified as:

Any solution that is not a^2 + b^2 = c^2 can be written as

a^n + b^n = c^n

and that can be rewritten as:

d^2 + e^2 = c^2 + k

So ANY solution to Fermat's Last Theorem HAS to be of the form:

d^2 + e^2 - k = c^2

If we add k back, we get a^n + b^n = c^n

For ANY solution.

k must be a multiple of c

k must raise both d and e by the same indice by multiplication BUT d and e cannot be the same number. Why?

That's very last and most obvious question to be the winner, take this and be the first to trivialise this problem. I'm not going to post, it can be yours, all you have to do is show why d and e cannot be the same number and you'll be a maths God, yawn.

 

[noelbcornerstone]

Because $k=c^2-a^2-b^2$ is so arbitrary that it is useless, furthermore, division by $c$ leaves the realm of integers. And what does it have to do with $a^3+b^3=c^3$?

Besides that, I have no idea what you are talking about. Please write down the case $n=4$ step by step, and you will see where the problems are; or I can tell you. But this general handwavy form is almost impossible to deal with.

 

[noelbcornerstone]

Ok, here's the bit to understand. a^2 + b^2 = c^2, we can take any integer d and write it in this form as we are for c and for some it is an integer, others it is not a whole number. This is indisputable.

Every other integer d, can be written in the form:
a^2 + b^2 - k = c^2 where is d is again presented by c^2

Do you understand that d is c^2 in this discussion?

It is the exact same statement as: a number is the sum of two squares or it is not the sum of two squares.

Then, for ANY solution we know it is of the form:

a^2 + b^2 - k = c^2 since this is the form any integer d takes that is not the sum of two squares, so c^2 in this instance is not the sum of two squares and represents any integer d that is not the sum of two squares.

ALL integers d are covered. If there is a d that is c^n and that is the sum of a^n plus b^n it is covered by the above. This is also indisputable. A number is either the sum of two squares or it isn't.

So, in the second equation, the variable k stores the additional indices of a and b for the solution if it is added, and this must also be a multiple of c, since the right hand side is a multiple of c, so if we add it, c's indice increase, as do a and b's

 

[fresh_42]

Repetition does not add value. Please write down the case $n=4.$

 

[noelbcornerstone]

For n = 4

a^n + b^n = c^4, a doesn't b doesn't c, all are integers

This can be rewrittens as:

d^2 + e^2 = c^2 where a^n and b^n are equal to the sum of two squares that is itself a square.

For ALL other values for c^4, we can rewrite as follows:

a^4 + b^4 = c^4

d^2 + e^2 - k = c^2, d != e

We took off the amount needed to make c^4 a square, we didn't divide by c

We took off the amount to drop of a and b by two indices.

k must be divisible by c or no solution for c^4 exists.

k must equal c^2 to exist, and d and e would have to be equal for a solution.

 

[fresh_42]

For n = 4

a^n + b^n = c^4, a doesn't b doesn't c, all are integers

This can be rewrittens as:

d^2 + e^2 = c^2 where a^n and b^n are equal to the sum of two squares that is itself a square.

For ALL other values for c^4, we can rewrite as follows:

a^4 + b^4 = c^4

d^2 + e^2 - k = c^2, d != e

We took off the amount needed to make c^4 a square, we didn't divide by c

We took off the amount to drop of a and b by two indices.

k must be divisible by c or no solution for c^4 exists.

k must equal c^2 to exist, and d and e would have to be equal for a solution.

This is nonsense.

 

[Al_]

Getting back to the first point, wouldn't it be amazing fun to discover a proof that Fermat could have found using the tools he knew? He had a good reputation at the time, and it would have been out of character for him to claim to have a proof that he did not have. Maybe there is a proof out there, waiting to be discovered by someone with a grasp of rather simple math and a lot of persistence!

 

[PeroK]

Getting back to the first point, wouldn't it be amazing ...

It would be amazing!

 

[doug618]

fib(n)^2 + Fib(n+1)^2 =Fib(2n+1) Fib = fibonacci number

fib(n)^3 +fib(n+1)^3 = fib(n-1)^3 + fib(3n+1)
Wouldn't it be great if this relationship between the fibonacci sequence and Fermat's theorem could be somehow used in a proof.

 

[Hornbein]

Getting back to the first point, wouldn't it be amazing fun to discover a proof that Fermat could have found using the tools he knew? He had a good reputation at the time, and it would have been out of character for him to claim to have a proof that he did not have. Maybe there is a proof out there, waiting to be discovered by someone with a grasp of rather simple math and a lot of persistence!

But he didn't really claim a proof. It was just a quick note in a book.

 

[doug618]

There are inifinitely many Pythagorean triplets ($n=2$) and no solution for $n=4$, why? Euler needed two articles to cover the case $n=3,$ which is a sure indicator that it is not trivial.

But even if we accept the case $n=3$ as given, there is no such thing as an induction. Any induction step fails.

fib(n)^2 + Fib(n+1)^2 =Fib(2n+1) Fib = fibonacci number
fib(n)^3 +fib(n+1)^3 = fib(n-1)^3 + fib(3n+1) Wouldn't it be great if this relationship between the fibonacci sequence and Fermat's theorem could be somehow used in a proof.

 

[pinball1970]

But he didn't really claim a proof. It was just a quick note in a book.

I thought he did? But it was too big to fit in the margin? (I read Simon Singh's book so I'm an expert.....)

 

[pinball1970]

"I have discovered a truly marvelous proof of this, which this margin is too small to contain."

 

[fresh_42]

I thought he did? But it was too big to fit in the margin? (I read Simon Singh's book so I'm an expert.....)

My article cites the exact wording (Latin) and its English translation:

“Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et generaliter nullam in infinitum ultra quadratum potestatem in duas ejusdem nominis fas est dividere: cujus rei demonstrationem mirabilem sane detexi. Hanc marginis exiguitas non caperet.”

“It is not possible, however, to decompose a cube into two cubes, or a biquadrate into two biquadrates, and in general, to decompose a power higher than the second into two powers with the same exponent: I have discovered a truly wonderful proof for this, but this margin is too narrow to grasp here.”

Source: https://www.physicsforums.com/insights/fermats-last-theorem/

 

[bob012345]

My article cites the exact wording (Latin) and its English translation:

“Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et generaliter nullam in infinitum ultra quadratum potestatem in duas ejusdem nominis fas est dividere: cujus rei demonstrationem mirabilem sane detexi. Hanc marginis exiguitas non caperet.”

“It is not possible, however, to decompose a cube into two cubes, or a biquadrate into two biquadrates, and in general, to decompose a power higher than the second into two powers with the same exponent: I have discovered a truly wonderful proof for this, but this margin is too narrow to grasp here.”

Source: https://www.physicsforums.com/insights/fermats-last-theorem/

Could we say his ‘proof’ wasn’t even marginal?

 

[fresh_42]

Could we say his ‘proof’ wasn’t even marginal?

It is hard to tell, since he didn't even wrote that in any letter as it was usual at his time. He only scribbled this into Diophant's book, and it was his son who published it after his death, sorting his remains.

I think he had a proof for $n=4,$ which relies on the argument for $n=2$ which he certainly knew, and possibly some ideas for $n=3,$ and he thought that these could be extended to any number. But this is an opinion. We do not have reliable information.

 

[elias001]

@fresh_42 there is an ongoing effort to try to formalise Wiles' proof in Lean. It would be perfect for anyone who likes to gatekeep on who or what (artifical silicon or carbon based life forms) qualify as reliable to explain mathematical concepts. See it is a bit late for these people to win the fields medal, but at least they can show their mathematical talent by being useful to the research math community. Their effort in helping out will be credited in the modern history of mathematics.

 

[martinbn]

@fresh_42 there is an ongoing effort to try to formalise Wiles' proof in Lean.

Can you say more about this. What does it mean to formalise? What is the goal? What is the motivation?

 

[elias001]

@martinbn the effort is lead by Kevin Buzzard. The goal is to use formalize Wiles proof by coding the entire proof using Lean 4's syntax (might be v3). The ultimate goal is to check whether the proof is correct in all its details by running it through a proof checker. Professor Buzzard was recruiting hrad students a year or so back and the requirements were that they have a grad course in algebraic number theory and have a working knowledge of the Lean programming language. (i will go look up the X/twitter post later on today). Professor Buzzard runs in the same circle online as others who are excited about what the latest in LLM coupled with Lean has to offer as assistance in math research. When I find the twitter post, i will reply it on here.

 

[elias001]

@martinbn

Here is a preprint paper.

By the way, I am not sure if you are on X/twitter. If you had or have not been for awhile and you feel like engaging with anyone in math on there, I should let you know that many in the online math community who are super allergic to the current US political climate has all migrated to Bluesky. Some of them has totally left and others have account on both X/Twitter and Bluesky. For those who has accounts on both sites, by transitivity, you can guess what they the political leanings are. Why am I telling you this. Just in case someone tries to roped into any kind of topics that are consider hot potatoes. (Any and all current ongoing geopolitical conflicts around the world)

@fresh_42 sorry did not mean to go off topic.

 

[fresh_42]

Such an endeavor comes with the usual problem: who checks the checkers? E.g., I had used a computer program to verify a theorem that used root systems, and I used it for the exceptional cases. Later on, I found a flaw. IIRC, I missed some cases. I was able to check them manually. Nevertheless, it shows that such an approach needs to be checked, too. And I think it is more reasonable to learn the math than learning the syntax of the checker. The 4-color theorem was the first big case of a theorem that used computer power. And the first version had a flaw. Wiles's first version of FLT also had a flaw, which he was able to fix with the help of Richard Taylor. Checking proofs requires dialogue. Otherwise, we replace one black box with another without creating trust. Furthermore, if I learn the language Wiles had used, I will study mathematics. If I learn Lean what the AI did, I will waste my time on learning an otherwise useless language. So wouldn't it make more sense to investigate efforts in studying the first black box?

This case here sounds like a nice project, but it is otherwise useless. The checker needs to be checked, and to enable checking the checker requires building useless skills. I'd rather build up the skill to understand the original paper. We trusted Wiles's proof because some (more than one) human mathematicians had checked it. Why should we trust a machine, or even more, its coding?

 

[elias001]

@fresh_42 You should ask Terry Tao, Peter Scholze, and Tim Gowers for their opinions. Tao has used it in his online blog post. Sholze asked Prof Buzzard awhile back to check whether one of his proofs to some number theory problem is correct. He asked people working in that community help with formalising the proof in Lean's syntax. As for checking the checkers, I think you should ask peeps working in formal verifications in computer science. Some and at least one of them (who is also on X) is familiar with what is aware with the happenings of the Math & Lean project. My very basic understanding are that both communities, the ones eho does math proof checking and formal verification of software has a common language between the two disciplines. Oh I should add, AI/LLM is not involved. I think there might be plans in bringing the two together. Also, Lean has been used in introduction to mathematical courses. I mean students learn how to program in Lean and learn to translate math statements and their own proof statements in Lean syntax. Lean checks if their proofs are correct.

 

[martinbn]

@martinbn the effort is lead by Kevin Buzzard. The goal is to use formalize Wiles proof by coding the entire proof using Lean 4's syntax (might be v3).

Buzzard of course is credible.

The ultimate goal is to check whether the proof is correct in all its details by running it through a proof checker.

I don't think so. Buzzard does not doubt that proof is correct, he is a student of Taylor and does understand the proof. He seems interested in formalising the proof for the sake of formalising it, not to check if it is correct.

 

[elias001]

@fresh_42 if memory serves me right, there was an article in Quanta magazine where Dusa McDuff was interviewed about the foundations of Sympletic geometry rests not on firm grounding. Sometimes later, Kevin Buzzard gave a talk about Lean and its role as a proof checker, mentioned the Quanta magazine's article, and how often times mistakes are found in journal articles, and what is done about them. There are of course pushbacks, Steven Krantz was one of them i think.

Here is Andrew Granville about Lean's role in Mathematics

Tao being interviewed on Lean here, and with copilot's help and without LLM's help.

 

[elias001]

@martinbn @fresh_42 Here was Buzzzard's talk that is referred to in my previous reply. And here is his recent talk on formalizing Fermat in Lean.

 

[elias001]

@martinbn I think for everyone else's and their prosterity's sake and their peace of mind, that every little details however minor are checked out.

 

[fresh_42]

I said it is a nice project. It means that students must learn the proof. What it does not mean is that we have a reliable proof of proof. It doesn't add value for anyone outside the project. In the late 80s, some people coded an OS written in BASIC on a mainframe. So? It wasn't meant to create another OS; it was a project to improve coding skills. This is the same in this case. The project does not replace anything; it teaches the students a proof and coding skills.